# Need help on easy differential equation

• schattenjaeger
In summary, you figure multiply out the dt, so dx=(.8*sqrt(100+x^2)/x dt, then integrate so x=allthatstuff *t, except I'm not sure that's right, should everything be constant like that?\frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x} In summary, you have:\frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x}\int \frac{4}{25}\,dt = \int \frac{x}{\

#### schattenjaeger

I actually haven't had differential equations, but I managed to somehow to register for the class I am in-_-, fortunately it's easy, but I'm still not sure how

dx/dt = (.8*sqrt(100+x^2)/x

I figure multiply out the dt, so dx=(.8*sqrt(100+x^2)/x dt, then integrate

so x=allthatstuff *t, except I'm not sure that's right, should everything be constant like that?

Last edited:
$$\frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x}$$

I THINK you subtract the RHS so then its off the form:

$$Mx' + Nx = 0$$

I didnt take diff eq either and don't really know much about it, but if you can take it from there, go for it. My next thought would be some kind of integrating factor.

What do you know about separation of variables...?

Daniel.

http://pacific.uta.edu/~qiming/Project2.htm [Broken]

There's the actual problem, I already did the computational part and all that, now I just need to compare my result with the analytic one, which I don't know how to do:(

Last edited by a moderator:
It's a simple integration.

Can u compute

$$\int \frac{\sqrt{x^{2}+100}}{x} \ dx$$...?

U could use a hyperbolic substitution

$$x=10\sinh t$$

Daniel.

err, how did you get that? I wound up with x on top(in which case you can use the sub u=100+x^2

but I haven't had the class yet so I've really got no clue

EDIT: My chief concern is no matter what way I try to solve it, I'm not getting a result that really makes sense with the given problem, and certainly doesn't match my numerical answer.

Last edited:
You have:

$$\frac{dx}{dt} = \frac{4}{5}\frac{\sqrt{100+x^2}}{5x}$$

Therefore:

$$\int \frac{4}{25}\,dt = \int \frac{x}{\sqrt{100+x^2}} \, dx + c$$

To do the integral on the right hand side, put $u = 100 + x^2$. Then

$$du = 2x dx$$

and the integral becomes

$$\int \frac{1}{2\sqrt{u}} \, du$$

CAn you do it from there?

Yah, that's not the problem anymore

it's 5/4 sqrt(100+x^2) = t, the problem is look at the actual question in that link I posted. The answer doesn't seem to make sense. At least I don't think it does. Well, it doesn't match my numerical answer at least, which means I probably screwed up. *runs off to fix*

well yah, it doesn't make any sense considering the actual question

Last edited:
Yes,alright,it's that integral written by James R.Mine would have been a little tricky...

Daniel.

## 1. What is a differential equation?

A differential equation is a mathematical equation that describes the relationship between a function and its derivatives. It involves one or more variables and their rates of change.

## 2. Why are differential equations important?

Differential equations are used to model and solve a wide range of real-world problems in fields such as physics, engineering, economics, and biology. They are also fundamental in understanding and predicting natural phenomena.

## 3. How do I solve a differential equation?

The method for solving a differential equation depends on its type and complexity. Some common techniques include separation of variables, substitution, and using integrating factors. It is important to understand the problem and choose the appropriate method.

## 4. What is an easy differential equation?

An easy differential equation is one that has a simple form and can be solved using basic techniques. These equations typically involve only one or two variables and have a small number of derivatives.

## 5. Can I use a computer to solve a differential equation?

Yes, there are many software programs and online tools available that can solve differential equations numerically. However, it is still important to have a good understanding of the problem and the steps involved in solving it.

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